Calculate the equilibrium concentration of Zn

^{2+} in a Zn(CN)

_{4}^{2-}My Answer:-

Zn

^{2+}(aq)+ 4CN (aq)

ZnCN

_{4}^{2-}(aq) ; K

_{f }=1×10

^{18}Let x be the change in concentration as Zn

^{2+} dissociates. Because the initial Zn

^{2+} concentration is 0, the concentration at any times is x:

[tex]1.0\times10^{18}=\frac{[Zn(CN)_4]^{2-}}{[Zn^{2+}][CN^-]^4}=\frac{0.30-x}{x(4x)^4}[/tex]

[tex]1\times 10^{18}*(256x^5)=0.30- x[/tex]

Since x is very small in comparison to 0.30 M, drop x:

Solving this equation, we get x=6.51 × 10

^{-5} M

Is this answer correct?